# How do you find a unit vector normal to the surface x^3+y^3+3xyz=3 ay the point(1,2,-1)?

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23
Oct 25, 2016

$\left\{- \frac{1}{\sqrt{14}} , \frac{3}{\sqrt{14}} , \sqrt{\frac{2}{7}}\right\}$

#### Explanation:

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$f \left(x , y , z\right) = {x}^{3} + {y}^{3} + 3 x y z - 3 = 0$

The gradient of $f \left(x , y , z\right)$ at point $x , y , z$ is a vector normal to the surface at this point.

The gradient is obtained as follows

$\nabla f \left(x , y , z\right) = \left({f}_{x} , {f}_{y} , {f}_{z}\right) = 3 \left({x}^{2} + y z , {y}^{2} + x z , x y\right)$ at point
$\left(1 , 2 , - 1\right)$ has the value
$3 \left(- 1 , 3 , 2\right)$ and the unit vector is
$\frac{\left\{- 1 , 3 , 2\right\}}{\sqrt{1 + {3}^{2} + {2}^{2}}} = \left\{- \frac{1}{\sqrt{14}} , \frac{3}{\sqrt{14}} , \sqrt{\frac{2}{7}}\right\}$

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